Question

solve y-y1=m(x-x1) for x1.

Answer 1

Distribute

Answer 2

y - y1 = mx - mx1

mx1 = mx - y + y1

x1 = (mx - y + y1)/m

Answer 3

!.Divide both sides by m
2.Subtract x both sides.
3.Divide by -1 both sides

Answer 4

How to solve this problem " passes through (-4,2) parrallel to the line whose equation is y=1/2x+5 ??

Answer 5

For parallel lines the gradients are equal.
So the gradient of the required line=1/2
\(y-y_1=m(x-x_1)\)
\(y-2=\frac{1}{2}(x-(-4)\)
\(y-2=\frac{1}{2}(x+4)\)
solve for y.
Can u do?

Answer 6

No

Answer 7

\(y-2=\frac{1}{2}(x+4)\)
\(y-2=\large\frac{x}{2}+\frac{4}{2}\)
\(y-2=\large\frac{x}{2}+2\)
\(y=\large\frac{x}{2}+2+2\)
\(y=\large\frac{x}{2}+4\)
Is it clear?

Answer 8

Why does the 1 in the fraction turn to x?

Answer 9

When u multiply 1 by x u get x. Is that clear?

Answer 10

Yes

Answer 11

Yes

Answer 12

ok 5yn:)

Answer 13

What If its passes through (1,-1) parrallel to the line that passes through (4,1) and (2,-3)

Answer 14

plz post a question per post.

Answer 15

1.Find the gradient \(m_1\)of the line passing through (4,1) and (2,-3) usin the formula
\[m_1=\frac{ y_2-y_1 }{ x_2-x_1 }\]
\[=\frac{ -3-1 }{ 2-4 }\]
2.For parallel lines gradients are equal.
\(m=m_1\)
so
\[m_1=\frac{ -3-1 }{ 2-4 }\]
\[\frac{ -3-1 }{ 2-4 }=\frac{ y-y_1 }{ x-x_1}\]
\[\frac{ -3-1 }{ 2-4 }=\frac{ y-(-1) }{ x-1}\]

Answer 16

Then it turns to 12/2 right??

Answer 17

\[\frac{ -3-1 }{ 2-4}=\frac{ y+1 }{ x-1 }\]
\[\frac{ -4 }{ -2 }=\frac{ y+1 }{ x-1}\]
\[2=\frac{ y+1 }{ x-1}\]

Answer 18

Oh ok:)